Does anyone know the equation to this trigonometric function? Step by step?

A general cosine function (we could also use a sine function) is written as:

y = A*cos(k*x + p) + M

We will find that the function of the graph is:

f(x) = 2*cos(2*x + 2.09) - 2

Let's return to the general function:

y = A*cos(k*x + p) + M

A is the amplitude, it defines the distance between the value of a maximum and the value of the minimum, such that A is exactly half of that difference.

Here we can see that the maximum is 0, and the minimum is -4

The differene is: 0 - (-4) = 4

Then:

A = 4/2 = 2

f(x) = 2*cos(k*x + p) + M.

M is the midline, this is, the horizontal line that cuts the graph in two halves. Here we can see that the midline is x = -2, then:

M = -2

f(x) = 2*cos(k*x + p) - 2

p is the phase shift.

In the graph, we can see that f(0) = -3, so we have:

f(0) = 2*cos(0 + p) - 2 = -3

      cos(p) = -1/2

             p = Acos(-1/2) = 2.09

Then we have:

f(x) = 2*cos(k*x + 2.09) - 2

Finally, k is related to the frequency of the function.

We can see that the function does a complete cycle at x = pi

This means that:

f(x) = f(x + pi)

Knowing that the period of a cosine function is 2*pi, then:

k*(x + pi) = k*x + 2*pi

k = 2

Then the equation of the graph is:

f(x) = 2*cos(2*x + 2.09) - 2

If you want to learn more, you can read:

https://brainly.com/question/24372261

Answer(s):

[tex]\displaystyle y = 2sin\:(2x + 1\frac{1}{4}\pi) - 2 \\ y = 2cos\:(2x - 1\frac{1}{4}\pi) - 2[/tex]

Step-by-step explanation:

[tex]\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow -2 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-\frac{5}{8}\pi} \hookrightarrow \frac{-1\frac{1}{4}\pi}{2} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 2[/tex]

OR

[tex]\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow -2 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\frac{5}{8}\pi} \hookrightarrow \frac{1\frac{1}{4}\pi}{2} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 2[/tex]

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of sine, then by all means, go for it, but be careful and follow what is explained here. Now, as you can see, the photograph on the right displays the trigonometric graph of [tex]\displaystyle y = 2sin\:(2x - 1\frac{1}{4}\pi) - 2,[/tex] in which you need to replase “cosine” with “sine”, then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosine graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the sine graph [photograph on the right] is shifted [tex]\displaystyle \frac{\pi}{4}\:unit[/tex] to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACKWARD [tex]\displaystyle \frac{\pi}{4}\:unit,[/tex] which means the C-term will be negative. Now, BEFORE we go any further, we must remember that this particular cosine graph [thank goodness it is a cosine graph we are working with] ALREADY has a horisontal shift and does not have a single crest oscıllαtıng about any endpoint on the y-axis. So, in this case we need to figure out how far the FIRST oscıllαtıng crest is from the origin, and that obviously would be [tex]\displaystyle \frac{5}{8}\pi\:units.[/tex] Though, sinse we want the sine equation of this graph, it must be “negative”; so, by perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{-\frac{5}{8}\pi} = \frac{-1\frac{1}{4}\pi}{2},[/tex] in which the value of C is [tex]\displaystyle -1\frac{1}{4}\pi.[/tex] So, the sine equation of the cosine graph, accourding to the horisontal shift, is [tex]\displaystyle y = 2sin\:(2x + 1\frac{1}{4}\pi) - 2.[/tex] Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits [tex]\displaystyle [\frac{7}{8}\pi, -2],[/tex] from there to [tex]\displaystyle [-\frac{\pi}{8}, -2],[/tex] they are obviously [tex]\displaystyle \pi\:units[/tex] apart, telling you that the period of the graph is [tex]\displaystyle \pi.[/tex] Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at [tex]\displaystyle y = -2,[/tex] in which each crest is extended two units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

**As you can see, this is one of those moments where you will really need to be careful because if you notised, both equations have OPPOCITE horisontal shifts and C-values. Now, the ONLY TIME this occurs is when all crests in a SINUSOIDAL graph cycle half-way in between endpoints. Your best bet is to jot this down for when you see graphs like these in the future.

I am delighted to assist you at any time.

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